.
Recall &epsilonmach
is a very small number (compared to 1).
1.2
X2. (a) [3 points] The functional iteration you did in HW1
with g(x) = cos(x) did converge to a fixed point of g,
i.e. to a root of f(x) = cos(x) - x.
Evaluate g'(r) for that case and explain how this is
consistent with Theorem 1.6.
(b) [1 point] The equation 4*x*(1-x) = x has a positive
solution. What is it?
(c) [2 points] If you couldn't do (b) analytically,
and tried iterating the function
g(x) = 4*x*(1-x) to try to find the solution numerically,
explain why Theorem 1.6 says you will be unsuccessful.
(d) [2 points] Run this code,
hw2_ringland.m, and describe what actually happens when you iterate the
function g in (b), starting with x0 = 0.4, for example.
(e) [1 point] What does the line starting "ia =" do?
Note that the semi-colon (;) at the end of a line in
Octave or Matlab suppresses the output. Remove the semi-colon to see the value
of the evaluated expression.
(f) [1 point] What do the single-quotes (') do in the 3rd line from the bottom?
(g) [1 point] Observe that a PostScript graphics file has been produced by
running the Octave code. How do you (i) view, (ii) print this file
on the computer/operating system you're using?
1.4
2 (a) [4 points] (A few steps of Newton's method by hand.)
X3. GRADS ONLY, BUT UNDERGRADS CAN DO FOR EXTRA CREDIT.
( (a) 3 points, (b) 7 points )
(From Atkinson.) We want to create a method for calculating the square
root
of any (nonnegative) number, a, that uses a fixed
number of Newton iterations.
We can write
a = b * 2m
where m is an even integer and 1/4 <= b < 1. Then
sqrt(a) = sqrt(b) * 2m/2, 1/2 <= sqrt(b) < 1 .
Thus the problem reduces to calculating sqrt(b) for 1/4 <= b < 1.
Use the linear interpolating formula
x0 = (2b+1)/3 for 1/4 <= b < 1
as an initial guess for Newton iteration to find sqrt(b).
(a) Bound the error sqrt(b)-x0.
(b) Estimate how many iterates are necessary in order that
0 <= | xi - sqrt(b) | <= 2-52
which is about the limit of IEEE double precision.
Hint for (b): Let f(x) = x2 - b.
Write Taylor's theorem with the remainder at quadratic order,
expanding at xi and evaluating at sqrt(b). I.e.
f(sqrt(b)) (which = 0 ) = f(xi) + (sqrt(b)-xi)
f'(xi) + R2.
Use this to get a relationship between the error at step i and the
error at step i+1.